Result and proof on the conditional expectation of the product of two random variables

Solution 1:

No, in general the statement is not correct.

Example Consider sets $A_1,A_2,A_3$ such that $\mathbb{P}(A_1 \cap A_2 \cap A_3) = 0$ and $\mathbb{P}(A_i)>0$, $\mathbb{P}(A_i \cap A_j) = \mathbb{P}(A_i) \cdot \mathbb{P}(A_j)$ for any two $i,j \in \{1,2,3\}$. Set $X:=1_{A_1}$, $Y=1_{A_2}$, $\mathcal{F} := \sigma(1_{A_3})$. Then $X$ and $Y$ are independent, as well as $X$ and $\mathcal{F}$, so the assumption on the independence is satisfied. For $F:= A_3$ we have

$$\int_F X \cdot Y \, d\mathbb{P} = \int 1_{A_1} \cdot 1_{A_2} \cdot 1_{A_3} \, d\mathbb{P} = \mathbb{P}(A_1 \cap A_2 \cap A_3) = 0 \tag{1} $$

On the other hand,

$$\int_F \mathbb{E}(X) \cdot \mathbb{E}(Y \mid \mathcal{F})\, d\mathbb{P} = \mathbb{P}(A_1) \cdot \underbrace{\mathbb{E}(Y)}_{\mathbb{P}(A_2)} \cdot \mathbb{P}(A_3)>0 \tag{2}$$

where we used that $Y$ and $\mathcal{F}$ are independent (hence $\mathbb{E}(Y \mid \mathcal{F})=\mathbb{E}(Y)$). Since by the definition of conditional expectation

$$0 \stackrel{(1)}{=} \int_F X \cdot Y \, d\mathbb{P} \stackrel{!}{=} \int_F \mathbb{E}(X \cdot Y \mid \mathcal{F}) \, d\mathbb{P}$$

we conclude from $(2)$ that

$$\mathbb{E}(X \cdot Y \mid \mathcal{F}) \neq \mathbb{E}(X) \cdot \mathbb{E}(Y \mid \mathcal{F})$$

But: If $\sigma(X)$ is independent of $\sigma(Y,\mathcal{F})$ (i.e. the smallest $\sigma$-algebra containing $\mathcal{F}$ and $\sigma(Y)$), then the claim holds as the following proof shows:

Denote by $\mathcal{G} := \sigma(\mathcal{F},\sigma(Y))$ the $\sigma$-algebra generated by $\mathcal{F}$ and $Y$. Since $\sigma(X)$ and $\mathcal{G}$ are independent, we have

$$\mathbb{E}(X \mid \mathcal{G}) = \mathbb{E}(X)$$

Consequently, by the tower property

$$\mathbb{E}(X \cdot Y \mid \mathcal{F}) = \mathbb{E}(Y \cdot \underbrace{\mathbb{E}(X \mid \mathcal{G})}_{\mathbb{E}(X)} \mid \mathcal{F}) = \mathbb{E}(X) \cdot \mathbb{E}(Y \mid \mathcal{F})$$